使用foldr Haskell反转列表

Question

我正在尝试使用foldr来反转Haskell中的列表列表。有一个我想做的例子：

> reverse'' [[1,2,3],[3,4,5]]
[[5,4,3],[3,2,1]]

我的代码（它没有用）：

reverse'':: [[a]]->[[a]]
reverse'' x = foldr (\n acum -> acum:(foldr(\m acum1 -> acum1++[m])) [] n) [[]] x

我的IDE在第二个折叠开始时报告错误。

Answer 1

为了分析你当前的解决方案，我已经将你的单行分解为一些辅助功能，其类型根据其组成推断：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [[]] x
    where
        f :: [a] -> a -> [a]
        f n acum = acum : reverse n

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

当我尝试编译上面的内容时，ghc对reverse''的类型签名抱怨以下错误：

Expected type: [[a]] -> [[a]] -> [[a]]
  Actual type: [[a]] ->  [a]  -> [[a]]

我做了一些挖掘，为了使reverse''具有[[a]] -> [[a]]类型，函数f需要具有类型[a] -> [[a]] -> [[a]]。但是，当前的f类型为[a] -> a -> [a]，在这种情况下为[[a]] -> [a] -> [[a]]。

以下具有正确的类型，但由于累加器的起始值，在数组的开头插入额外的[]值：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [[]] x
    where
        f :: [a] -> [[a]] -> [[a]]
        f n acum = acum ++ [reverse n]

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

最终解决方案

通过将初始累加器值更改为空列表[]，而不是空列表[[]]的列表，我们最终得到了一个可行的解决方案：

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [] x
    where
        f :: a -> [a] -> [a]
        f n acum = acum ++ [reverse n]

        reverse :: [a] -> [a]
        reverse n = foldr append [] n

        append :: a -> [a] -> [a]
        append m acum = acum ++ [m]

作为单行

如果你真的想要这个作为一个单行，那么它是：

reverse'' :: [[a]] -> [[a]]
reverse'' = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) []

工作：

append m acum = acum ++ [m]
append = \m acum1 -> acum1 ++ [m]

reverse n = foldr append [] n
reverse = \n -> foldr append [] n
reverse = foldr append []
reverse = foldr (\m acum1 -> acum1 ++ [m]) []

f n acum = acum ++ [reverse n]
f = \n acum -> acum ++ [reverse n]
f = \n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]

reverse'':: [[a]] -> [[a]]
reverse'' x = foldr f [] x
reverse'' x = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) [] x
reverse'' = foldr (\n acum -> acum ++ [foldr (\m acum1 -> acum1 ++ [m]) [] n]) []

附录：明确的递归解决方案

这是一个明确的递归解决方案（即不使用fold），提供了map和reverse的定义。

reverse :: [a] -> [a]
reverse (x:xs) = reverse xs ++ [x]
reverse []     = []

map :: (a -> b) -> [a] -> [b]
map f (x:xs) = f x : map f xs
map _ [] = []

reverse'' :: [[a]] -> [[a]]
reverse'' ls = reverse $ map reverse ls